Solve the equation `sin^(2n-1)x+2cos^(2n-1)x=2`,where `n in N`.

The general solution of the equation sin^(2)x + cos^(2)3x =1 is equal to : (where n in I )

The solution of the equation sin^(3)x+sin x cos x+cos^(3)x=1 is (n in N)2n pi(b)(4n+1)(pi)/(2)(2n+1)pi(d) None of these

Statement -1 the solution of the equation (sin x)/(cos x + cos 2x)=0 is x=n pi, n in I Statement -2 : The solution of the equation sin x = sin alpha , alpha in [(-pi)/(2),(pi)/(2)] x={npi+(-1)^(n)alpha,n in I}

Evaluate: quad int_(-(pi)/(2))^((pi)/(2))sqrt(cos^(2n-1)x-cos^(2n+1)x)xdx, where n in N

Solve for x the equation |(a^(2),a,1),(sin(n+1)x,sin nx,sin(n-1)x),(cos(n+1)x,cosn x,cos(n-1)x)|=0

Prove that: sin(n+1)x sin(n+2)x+cos(n+1)x cos(n+2)x=c

If n in N and the set of equations,(sin^(-1)y)^(2)+(cos^(-1)x)=(n pi^(2))/(4) and (sin^(-1)y)^(2)-(cos^(-1)x)=(pi^(2))/(16) is consistent,then n can be equal to-

Show that the roots of the equation (1+x)^(2n)+(1-x)^(2n)=0 are given by x=+-i tan{(2k-1)(pi)/(2n)) where k=1,2,3,...n

The equatin 2x = (2n +1)pi (1 - cos x) , (where n is a positive integer)